A particle moves in xy plane with constant acceleration a directed along negative y-axis.The equation of path of the particle has form y=bx-cx^2,where a and b are positive constants.Find the velocity of the particle at the origin of coordinates.ANS:under root[a(1+b^2)/2c]REQUEST:Please...submit with sufficient explanation.
Sameer Gupta , 12 Years ago
Grade 12th Pass
2 Answers
Akash Kumar Dutta
Last Activity: 12 Years ago
Dear Sammer, Comparing the eq. with projectile from earth=>y=x.tan m -x^2tan m/R...where R=Range We get a=g...b=tan m.....c=tan m/R hence c=b/R=>R=b/c But R=u^2sin 2m / a Hence u^2=ab/2c(sin m).(cos m) as tan m=b,putting the value of sin m and cos m the eq becomes u^2=a(1+b^2)/2c Hence u=root[a(1+b^2)/2c] ANS. Regards.
Tushar
Last Activity: 5 Years ago
Y=bX-cX^2
Differentiating it
( V(y)=bV(x)-2cXV(x). )-#1
At origin X=0
So.( V(y)=bV(x). )-#2
Differentiating #1 further
a(y)=ba(x)-2c(V(x))^2-2ca(y)X
At origin X=0 and acc. Is along -ve Y so a(x) =0
-a(y)=a(total)
a=2c(V(x))^2
So V(x)^2= a/2c
From #2. V(y)^2=b^2.a/2c
Total V = (. V(x)^2+V(y)^2. )^1/2
So V= ( a(p^2+1)/2q. )1/2
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